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Saturday, 22 March 2014

Clifford circuits for stabilizer codes

The initial state is of the form $\ket{\psi}\ket{0}\ket{+}$, where $\ket{\psi}$ is an arbitrary single qubit state to be encoded. Two stabilizers $M_1$ and $M_2$ acting on the three qubits can be associated to this initial state such that $\ket{0}$ is a $+1$ eigenstate of $M_1$ and $\ket{+}$ is a $+1$ eigenstate of $M_2$. Since $Z\ket{0}=\ket{0}$ and $X\ket{+}=\ket{+}$, then $M_1=IZI$ and $M_2=IIX$. The ``logical" Pauli operators in this case are simply $\overline{X}=XII$ and $\overline{Z}=ZII$.

When passing through some gate $U$ in an encoding circuit the stabilizer generators and logical Pauli operators after the action of the gate $U$ can be updated to new stabilizer generators $M'_1, M'_2$ and new logical Paulis $\overline{X}', \overline{Z}'$ given by conjugation by $U$:\[M'_1=UM_1U^\dagger , M'_2=UM_2U^\dagger, \overline{X}'=U\overline{X}U^\dagger, \overline{Z}'=U\overline{Z}U^\dagger.\]

Therefore, the stabilizer generators of $S$ are $M_1=-YZY$ and $M_2=-XXY$, and the logical Pauli operators corresponding to generators of $N(S)/S$ are $\overline{X}=IYX$ and $\overline{Z}=-XXY$.

Let $\delta$ be an $n\times n$ matrix and $S$ a $2n\times 2n$ matrix given as\[S=\begin{pmatrix}0 & \delta \\\delta & 0 \end{pmatrix},\]where here $0$ is an $n\times n$ matrix. By definition, $S$ is symplectic if and only if $J=S^TJS$, where \[J=\begin{pmatrix}0 &I \\I & 0 \end{pmatrix}.\]Then since\[\begin{pmatrix}0 &I \\I & 0 \end{pmatrix}=\begin{pmatrix}0 &\delta^T \\\delta^T & 0 \end{pmatrix}\begin{pmatrix}0 &I \\I & 0 \end{pmatrix}\begin{pmatrix}0 &\delta \\\delta & 0 \end{pmatrix}=\begin{pmatrix}0 &\delta^T\delta \\\delta^T\delta & 0 \end{pmatrix},\]in order for $S$ to be a symplectic matrix it must be the case that $I=\delta^T\delta$. Let the columns of the matrix $\delta$ be given by $\vec{r}_i\in\mathbb{Z}^n_2$ for $i\in\{1,\dots,n\}$. Then the condition $I=\delta^T\delta$ can then be interpreted as requiring each row $\vec{r}_i$ to have unit norm, $\vec{r}_i\cdot\vec{r}_i=1 (mod \ 2)$, and also that distinct rows be orthogonal, $\vec{r}_i\cdot\vec{r}_j=0 (mod \ 2)$ for $i\neq j$.

Then the condition $\delta^T\delta=I$, implies that $\delta^T\delta=\delta^2=(\sum_{i=1}^{n}d_{a,i}d_{i,b})_{a,b}=(\delta_{a,b})_{a,b}=I$. Therefore, $\sum_{i=1}^{n}d_{a,i}d_{i,a}=\sum_{i=1}^{n}d^2_{a,i}=n-1(mod 2)=1$ and $\sum_{i=1}^{n}d_{a,i}d_{i,b}=n-2(mod 2)=0$, which implies that $n$ must be even. Hence the matrix \[S=\begin{pmatrix}0 & \delta \\\delta & 0 \end{pmatrix},\]is symplectic if and only if $n$ is even.

Recalling that $H^\dagger=H$ and $CNOT^\dagger=CNOT$, and reversing the order of the operations used above then yields a sequence of operations that implements $S$ (the rightmost operation is applied first) :